Question

New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April , ). Assume that room rates are normally distributed with a standard deviation of $55.

a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)?

b. What is the probability that a hotel room costs less than $140 per night (to 4 decimals)?

c. What is the probability that a hotel room costs between $200 and $300 per night (to 4 decimals)?

d. What is the cost of the 20% most expensive hotel rooms in New York City? Round up to the next dollar.

Answer 1

Solution :

Given that,

mean = = 204

standard deviation = =55

a ) P (x > 225 )

= 1 - P (x < 225 )

= 1 - P ( x -  / ) < ( 225 - 204/ 55)

= 1 - P ( z < 21 / 55)

= 1 - P ( z < 0.38 )

Using z table

= 1 - 0.6487

= 0.3513

Probability = 0.3513

b ) P( x < 140 )

P ( x - / ) < ( 140 - 204/ 55)

P ( z < - 64 / 55 )

P ( z < -1.16)

= 0.1223

Probability = 0.1223

c ) P (200 < x < 300 )

P ( 200 - 204/ 55) < ( x -  / ) < ( 300 - 204/ 55)

P ( - 4 / 55 < z < 96 / 55 )

P (-0.07 < z < 1.74 )

P ( z <1.74 ) - P ( z < - 0.07)

Using z table

= 0.9595 - 0.4710

= 0.4885

Probability = 0.4885

d ) P( Z z) =20%
P(Z z) = 0.20

z = - 0.84

Using z-score formula,

x = z * +

x = - 0.84 *55 + 204

x = 157.8

x = 157.8